# Chapter 3 - The Derivative - 3.3 Rates of Change - 3.3 Exercises - Page 158: 16

-16

#### Work Step by Step

For t = 2 the instantaneous rate of change is $\lim\limits_{h \to 0} \frac{s(2 + h) - s(2)}{h}$ $s(2+h)=-4(2+h)^{2} -6 =-4(4+4h + h^{2})-6=-4h^{2}-16h-22$ $s(2) = -22$ The instantaneous rate of change is then: $\lim\limits_{h \to 0} \frac{s(2 + h) - s(2)}{h} = \lim\limits_{h \to 0} \frac{-4h^{2}-16h-22+22}{h}=\lim\limits_{h \to 0} \frac{h(-4h-16)}{h} = -16$

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